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what is the solution of 5|x+1|+6=21

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how would we start it?
5|χ+1|=15 => |x+1| = 3.
im sure as long as you just get an answer; you will continue to type in the same kind of problem over and over again :)

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Other answers:

assuming x is real: \[5 \sqrt((x+1)^2)+6 = 21\] otherwise: |x+1|=3
don't just give it to him.
i don't know how to start it absolute value i just don't understand it
the absolute value is the last thing to work on; how do you get that absolute value all by itself?
5(x+1)+6=21 (5x+5)+6=21 5x+6=16 5x=10 x=2
looking at this question, what is the other solution for x if it was not an absolute value?
i'm seriously terrible in math i really don't know:[
the term "absolute value" means that there is a quantifiable value; something that can be measured. length, height, area, volume, distance are all examples of abslute values
dont jus give him the answer he is probably usin this site for a summer school!!!
in other words; an absolute value doesnt care about what direction the sign takes us in; it only cares about a quantifiable measure
|a| is always positive, you discount the negative answer.
sky, you isolate the absolute value just like you would isolate a variable. Llort, no you do not 'discount' the negative answer. It is a valid solution.
|-12| has the same absolute value as |12|
as a result; if we are determining any value for inside of the absolute value notation; we have to account for the positive and the negative results that occur within it
So if you have \[5|x+1| + 6 = 21\] You want to isolate the term with the absolute value in it first. You do this by subtracting 6 from both sides of the equation.
What do you get when you subtract the 6?
No, \(5|x+1| + 6 - 6 = 21 -6\) \[\implies 5|x+1| + 0 = 15\]\[\implies 5|x+1| = 15\]
So now lets say that |x+1| is something, lets call it k.. So we have \(5k = 15 \)\[\implies \frac{1}{5}\cdot5k = \frac{1}{5}15\]\[\implies k = 3\] And since we said k was |x+1| we have \[|x+1| = 3\]So what values can x have and still make this equation true?
I am assuming the other answer is x=-4
x =-4 is one solution yes, because |-4 + 1 | = |-3| = 3

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